Calculus Examples

$\int_{0}^{\frac{π}{2}} \frac{\sin (x)}{1 - \cos (x)} d x$

Step 1

Let $u = 1 - \cos (x)$ . Then $d u = \sin (x) d x$ , so $\frac{1}{\sin (x)} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = 1 - \cos (x)$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $1 - \cos (x)$ .

$\frac{d}{d x} [1 - \cos (x)]$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $1 - \cos (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- \cos (x)]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [- \cos (x)]$

Step 1.1.2.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- \cos (x)]$

Step 1.1.3

Evaluate $\frac{d}{d x} [- \cos (x)]$ .

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Step 1.1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \cos (x)$ with respect to $x$ is $- \frac{d}{d x} [\cos (x)]$ .

$0 - \frac{d}{d x} [\cos (x)]$

Step 1.1.3.2

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$0 - - \sin (x)$

Step 1.1.3.3

Multiply $- 1$ by $- 1$ .

$0 + 1 \sin (x)$

Step 1.1.3.4

Multiply $\sin (x)$ by $1$ .

$0 + \sin (x)$

Step 1.1.4

Add $0$ and $\sin (x)$ .

$\sin (x)$

Step 1.2

Substitute the lower limit in for $x$ in $u = 1 - \cos (x)$ .

$u_{lower} = 1 - \cos (0)$

Step 1.3

Simplify.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

The exact value of $\cos (0)$ is $1$ .

$u_{lower} = 1 - 1 \cdot 1$

Step 1.3.1.2

Multiply $- 1$ by $1$ .

$u_{lower} = 1 - 1$

Step 1.3.2

Subtract $1$ from $1$ .

$u_{lower} = 0$

Step 1.4

Substitute the upper limit in for $x$ in $u = 1 - \cos (x)$ .

$u_{upper} = 1 - \cos (\frac{π}{2})$

Step 1.5

Simplify.

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Step 1.5.1

Simplify each term.

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Step 1.5.1.1

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$u_{upper} = 1 - 0$

Step 1.5.1.2

Multiply $- 1$ by $0$ .

$u_{upper} = 1 + 0$

Step 1.5.2

Add $1$ and $0$ .

$u_{upper} = 1$

Step 1.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 0$

$u_{upper} = 1$

Step 1.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$\int_{0}^{1} \frac{1}{u} d u$

Step 2

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$\ln (| u |)]_{0}^{1}$

Step 3

Evaluate $\ln (| u |)$ at $1$ and at $0$ .

$\ln (| 1 |) - \ln (| 0 |)$

Step 4

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| 1 |}{| 0 |})$

Step 5

Simplify.

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Step 5.1

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$\ln (\frac{| 1 |}{0})$

Step 5.2

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\ln (\frac{| 1 |}{0})$

Step 6

The expression contains a division by $0$ . The expression is undefined.

Undefined